/*
 * This software is a cooperative product of The MathWorks and the National
 * Institute of Standards and Technology (NIST) which has been released to the
 * public domain. Neither The MathWorks nor NIST assumes any responsibility
 * whatsoever for its use by other parties, and makes no guarantees, expressed
 * or implied, about its quality, reliability, or any other characteristic.
 */

/*
 * SingularValueDecomposition.java
 * Copyright (C) 1999 The Mathworks and NIST
 *
 */

package weka.core.matrix;

import java.io.Serializable;

/**
 * Singular Value Decomposition.
 * <P>
 * For an m-by-n matrix A with m &gt;= n, the singular value decomposition is an
 * m-by-n orthogonal matrix U, an n-by-n diagonal matrix S, and an n-by-n
 * orthogonal matrix V so that A = U*S*V'.
 * <P>
 * The singular values, sigma[k] = S[k][k], are ordered so that sigma[0] &gt;=
 * sigma[1] &gt;= ... &gt;= sigma[n-1].
 * <P>
 * The singular value decompostion always exists, so the constructor will never
 * fail. The matrix condition number and the effective numerical rank can be
 * computed from this decomposition.
 * <p/>
 * Adapted from the
 * <a href="http://math.nist.gov/javanumerics/jama/" target="_blank">JAMA</a>
 * package.
 *
 * @author The Mathworks and NIST
 * @author Fracpete (fracpete at waikato dot ac dot nz)
 * @author eibe@cs.waikato.ac.nz
 * @version $Revision$
 */
public class SingularValueDecomposition implements Serializable {

    /** for serialization */
    private static final long serialVersionUID = -8738089610999867951L;

    /**
     * Arrays for internal storage of U and V.
     * 
     * @serial internal storage of U.
     * @serial internal storage of V.
     */
    private double[][] U, V;

    /**
     * Array for internal storage of singular values.
     * 
     * @serial internal storage of singular values.
     */
    private double[] s;

    /**
     * Row and column dimensions.
     * 
     * @serial row dimension.
     * @serial column dimension.
     */
    private int m, n;

    /**
     * Construct the singular value decomposition
     * 
     * @param Arg Rectangular matrix
     */
    public SingularValueDecomposition(Matrix Arg) {

        // Derived from LINPACK code.
        // Initialize.
        double[][] A = null;
        m = Arg.getRowDimension();
        n = Arg.getColumnDimension();

        /*
         * Apparently the failing cases are only a proper subset of (m<n), so let's not
         * throw error. Correct fix to come later? if (m<n) { throw new
         * IllegalArgumentException("Jama SVD only works for m >= n"); }
         */

        boolean usingTranspose = false;
        if (m < n) {

            // Use transpose and convert back at the end
            // Otherwise m < n case may yield incorrect results (see above comment)
            A = Arg.transpose().getArrayCopy();
            usingTranspose = true;
            int temp = m;
            m = n;
            n = temp;
        } else {
            A = Arg.getArrayCopy();
        }

        int nu = Math.min(m, n);
        s = new double[Math.min(m + 1, n)];
        U = new double[m][m];
        V = new double[n][n];
        double[] e = new double[n];
        double[] work = new double[m];
        boolean wantu = true;
        boolean wantv = true;

        // Reduce A to bidiagonal form, storing the diagonal elements
        // in s and the super-diagonal elements in e.

        int nct = Math.min(m - 1, n);
        int nrt = Math.max(0, Math.min(n - 2, m));
        for (int k = 0; k < Math.max(nct, nrt); k++) {
            if (k < nct) {

                // Compute the transformation for the k-th column and
                // place the k-th diagonal in s[k].
                // Compute 2-norm of k-th column without under/overflow.
                s[k] = 0;
                for (int i = k; i < m; i++) {
                    s[k] = Maths.hypot(s[k], A[i][k]);
                }
                if (s[k] != 0.0) {
                    if (A[k][k] < 0.0) {
                        s[k] = -s[k];
                    }
                    for (int i = k; i < m; i++) {
                        A[i][k] /= s[k];
                    }
                    A[k][k] += 1.0;
                }
                s[k] = -s[k];
            }
            for (int j = k + 1; j < n; j++) {
                if ((k < nct) & (s[k] != 0.0)) {

                    // Apply the transformation.

                    double t = 0;
                    for (int i = k; i < m; i++) {
                        t += A[i][k] * A[i][j];
                    }
                    t = -t / A[k][k];
                    for (int i = k; i < m; i++) {
                        A[i][j] += t * A[i][k];
                    }
                }

                // Place the k-th row of A into e for the
                // subsequent calculation of the row transformation.

                e[j] = A[k][j];
            }
            if (wantu & (k < nct)) {

                // Place the transformation in U for subsequent back
                // multiplication.

                for (int i = k; i < m; i++) {
                    U[i][k] = A[i][k];
                }
            }
            if (k < nrt) {

                // Compute the k-th row transformation and place the
                // k-th super-diagonal in e[k].
                // Compute 2-norm without under/overflow.
                e[k] = 0;
                for (int i = k + 1; i < n; i++) {
                    e[k] = Maths.hypot(e[k], e[i]);
                }
                if (e[k] != 0.0) {
                    if (e[k + 1] < 0.0) {
                        e[k] = -e[k];
                    }
                    for (int i = k + 1; i < n; i++) {
                        e[i] /= e[k];
                    }
                    e[k + 1] += 1.0;
                }
                e[k] = -e[k];
                if ((k + 1 < m) & (e[k] != 0.0)) {

                    // Apply the transformation.

                    for (int i = k + 1; i < m; i++) {
                        work[i] = 0.0;
                    }
                    for (int j = k + 1; j < n; j++) {
                        for (int i = k + 1; i < m; i++) {
                            work[i] += e[j] * A[i][j];
                        }
                    }
                    for (int j = k + 1; j < n; j++) {
                        double t = -e[j] / e[k + 1];
                        for (int i = k + 1; i < m; i++) {
                            A[i][j] += t * work[i];
                        }
                    }
                }
                if (wantv) {

                    // Place the transformation in V for subsequent
                    // back multiplication.

                    for (int i = k + 1; i < n; i++) {
                        V[i][k] = e[i];
                    }
                }
            }
        }

        // Set up the final bidiagonal matrix or order p.

        int p = Math.min(n, m + 1);
        if (nct < n) {
            s[nct] = A[nct][nct];
        }
        if (m < p) {
            s[p - 1] = 0.0;
        }
        if (nrt + 1 < p) {
            e[nrt] = A[nrt][p - 1];
        }
        e[p - 1] = 0.0;

        // If required, generate U.

        if (wantu) {
            for (int j = nct; j < nu; j++) {
                for (int i = 0; i < m; i++) {
                    U[i][j] = 0.0;
                }
                U[j][j] = 1.0;
            }
            for (int k = nct - 1; k >= 0; k--) {
                if (s[k] != 0.0) {
                    for (int j = k + 1; j < nu; j++) {
                        double t = 0;
                        for (int i = k; i < m; i++) {
                            t += U[i][k] * U[i][j];
                        }
                        t = -t / U[k][k];
                        for (int i = k; i < m; i++) {
                            U[i][j] += t * U[i][k];
                        }
                    }
                    for (int i = k; i < m; i++) {
                        U[i][k] = -U[i][k];
                    }
                    U[k][k] = 1.0 + U[k][k];
                    for (int i = 0; i < k - 1; i++) {
                        U[i][k] = 0.0;
                    }
                } else {
                    for (int i = 0; i < m; i++) {
                        U[i][k] = 0.0;
                    }
                    U[k][k] = 1.0;
                }
            }
        }

        // If required, generate V.

        if (wantv) {
            for (int k = n - 1; k >= 0; k--) {
                if ((k < nrt) & (e[k] != 0.0)) {
                    for (int j = k + 1; j < nu; j++) {
                        double t = 0;
                        for (int i = k + 1; i < n; i++) {
                            t += V[i][k] * V[i][j];
                        }
                        t = -t / V[k + 1][k];
                        for (int i = k + 1; i < n; i++) {
                            V[i][j] += t * V[i][k];
                        }
                    }
                }
                for (int i = 0; i < n; i++) {
                    V[i][k] = 0.0;
                }
                V[k][k] = 1.0;
            }
        }

        // Main iteration loop for the singular values.

        int pp = p - 1;
        int iter = 0;
        double eps = Math.pow(2.0, -52.0);
        double tiny = Math.pow(2.0, -966.0);
        while (p > 0) {
            int k, kase;

            // Here is where a test for too many iterations would go.

            // This section of the program inspects for
            // negligible elements in the s and e arrays. On
            // completion the variables kase and k are set as follows.

            // kase = 1 if s(p) and e[k-1] are negligible and k<p
            // kase = 2 if s(k) is negligible and k<p
            // kase = 3 if e[k-1] is negligible, k<p, and
            // s(k), ..., s(p) are not negligible (qr step).
            // kase = 4 if e(p-1) is negligible (convergence).

            for (k = p - 2; k >= -1; k--) {
                if (k == -1) {
                    break;
                }
                if (Math.abs(e[k]) <= tiny + eps * (Math.abs(s[k]) + Math.abs(s[k + 1]))) {
                    e[k] = 0.0;
                    break;
                }
            }
            if (k == p - 2) {
                kase = 4;
            } else {
                int ks;
                for (ks = p - 1; ks >= k; ks--) {
                    if (ks == k) {
                        break;
                    }
                    double t = (ks != p ? Math.abs(e[ks]) : 0.) + (ks != k + 1 ? Math.abs(e[ks - 1]) : 0.);
                    if (Math.abs(s[ks]) <= tiny + eps * t) {
                        s[ks] = 0.0;
                        break;
                    }
                }
                if (ks == k) {
                    kase = 3;
                } else if (ks == p - 1) {
                    kase = 1;
                } else {
                    kase = 2;
                    k = ks;
                }
            }
            k++;

            // Perform the task indicated by kase.

            switch (kase) {

            // Deflate negligible s(p).

            case 1: {
                double f = e[p - 2];
                e[p - 2] = 0.0;
                for (int j = p - 2; j >= k; j--) {
                    double t = Maths.hypot(s[j], f);
                    double cs = s[j] / t;
                    double sn = f / t;
                    s[j] = t;
                    if (j != k) {
                        f = -sn * e[j - 1];
                        e[j - 1] = cs * e[j - 1];
                    }
                    if (wantv) {
                        for (int i = 0; i < n; i++) {
                            t = cs * V[i][j] + sn * V[i][p - 1];
                            V[i][p - 1] = -sn * V[i][j] + cs * V[i][p - 1];
                            V[i][j] = t;
                        }
                    }
                }
            }
                break;

            // Split at negligible s(k).

            case 2: {
                double f = e[k - 1];
                e[k - 1] = 0.0;
                for (int j = k; j < p; j++) {
                    double t = Maths.hypot(s[j], f);
                    double cs = s[j] / t;
                    double sn = f / t;
                    s[j] = t;
                    f = -sn * e[j];
                    e[j] = cs * e[j];
                    if (wantu) {
                        for (int i = 0; i < m; i++) {
                            t = cs * U[i][j] + sn * U[i][k - 1];
                            U[i][k - 1] = -sn * U[i][j] + cs * U[i][k - 1];
                            U[i][j] = t;
                        }
                    }
                }
            }
                break;

            // Perform one qr step.

            case 3: {

                // Calculate the shift.

                double scale = Math.max(Math.max(Math.max(Math.max(Math.abs(s[p - 1]), Math.abs(s[p - 2])), Math.abs(e[p - 2])), Math.abs(s[k])), Math.abs(e[k]));
                double sp = s[p - 1] / scale;
                double spm1 = s[p - 2] / scale;
                double epm1 = e[p - 2] / scale;
                double sk = s[k] / scale;
                double ek = e[k] / scale;
                double b = ((spm1 + sp) * (spm1 - sp) + epm1 * epm1) / 2.0;
                double c = (sp * epm1) * (sp * epm1);
                double shift = 0.0;
                if ((b != 0.0) | (c != 0.0)) {
                    shift = Math.sqrt(b * b + c);
                    if (b < 0.0) {
                        shift = -shift;
                    }
                    shift = c / (b + shift);
                }
                double f = (sk + sp) * (sk - sp) + shift;
                double g = sk * ek;

                // Chase zeros.

                for (int j = k; j < p - 1; j++) {
                    double t = Maths.hypot(f, g);
                    double cs = f / t;
                    double sn = g / t;
                    if (j != k) {
                        e[j - 1] = t;
                    }
                    f = cs * s[j] + sn * e[j];
                    e[j] = cs * e[j] - sn * s[j];
                    g = sn * s[j + 1];
                    s[j + 1] = cs * s[j + 1];
                    if (wantv) {
                        for (int i = 0; i < n; i++) {
                            t = cs * V[i][j] + sn * V[i][j + 1];
                            V[i][j + 1] = -sn * V[i][j] + cs * V[i][j + 1];
                            V[i][j] = t;
                        }
                    }
                    t = Maths.hypot(f, g);
                    cs = f / t;
                    sn = g / t;
                    s[j] = t;
                    f = cs * e[j] + sn * s[j + 1];
                    s[j + 1] = -sn * e[j] + cs * s[j + 1];
                    g = sn * e[j + 1];
                    e[j + 1] = cs * e[j + 1];
                    if (wantu && (j < m - 1)) {
                        for (int i = 0; i < m; i++) {
                            t = cs * U[i][j] + sn * U[i][j + 1];
                            U[i][j + 1] = -sn * U[i][j] + cs * U[i][j + 1];
                            U[i][j] = t;
                        }
                    }
                }
                e[p - 2] = f;
                iter = iter + 1;
            }
                break;

            // Convergence.

            case 4: {

                // Make the singular values positive.

                if (s[k] <= 0.0) {
                    s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
                    if (wantv) {
                        for (int i = 0; i <= pp; i++) {
                            V[i][k] = -V[i][k];
                        }
                    }
                }

                // Order the singular values.

                while (k < pp) {
                    if (s[k] >= s[k + 1]) {
                        break;
                    }
                    double t = s[k];
                    s[k] = s[k + 1];
                    s[k + 1] = t;
                    if (wantv && (k < n - 1)) {
                        for (int i = 0; i < n; i++) {
                            t = V[i][k + 1];
                            V[i][k + 1] = V[i][k];
                            V[i][k] = t;
                        }
                    }
                    if (wantu && (k < m - 1)) {
                        for (int i = 0; i < m; i++) {
                            t = U[i][k + 1];
                            U[i][k + 1] = U[i][k];
                            U[i][k] = t;
                        }
                    }
                    k++;
                }
                iter = 0;
                p--;
            }
                break;
            }
        }

        if (usingTranspose) {
            int temp = m;
            m = n;
            n = temp;
            double[][] tempA = U;
            U = V;
            V = tempA;
        }
    }

    /**
     * Return the left singular vectors
     * 
     * @return U
     */
    public Matrix getU() {
        return new Matrix(U, m, m);
    }

    /**
     * Return the right singular vectors
     * 
     * @return V
     */
    public Matrix getV() {
        return new Matrix(V, n, n);
    }

    /**
     * Return the one-dimensional array of singular values
     * 
     * @return diagonal of S.
     */
    public double[] getSingularValues() {
        return s;
    }

    /**
     * Return the diagonal matrix of singular values
     * 
     * @return S
     */
    public Matrix getS() {
        Matrix X = new Matrix(m, n);
        double[][] S = X.getArray();
        for (int i = 0; i < Math.min(m, n); i++) {
            S[i][i] = this.s[i];
        }
        return X;
    }

    /**
     * Two norm
     * 
     * @return max(S)
     */
    public double norm2() {
        return s[0];
    }

    /**
     * Two norm condition number
     * 
     * @return max(S)/min(S)
     */
    public double cond() {
        return s[0] / s[Math.min(m, n) - 1];
    }

    /**
     * Effective numerical matrix rank
     * 
     * @return Number of nonnegligible singular values.
     */
    public int rank() {
        double eps = Math.pow(2.0, -52.0);
        double tol = Math.max(m, n) * s[0] * eps;
        int r = 0;
        for (int i = 0; i < s.length; i++) {
            if (s[i] > tol) {
                r++;
            }
        }
        return r;
    }

}
